Question

If $x=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\mathrm{and}y=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, show that $3x^2+4xy-3y^2=4+\frac{56}{3}\sqrt{10}$.

Answer 1

According to question,
$x=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\mathrm{and}y=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$$\begin{gathered} x=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2+2\sqrt{10}}{3} \\ =\frac{7+2\sqrt{10}}{3}...\left(1\right) \\ y=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{{\left(\sqrt{5}-\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2-2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2-2\sqrt{10}}{3} \\ =\frac{7-2\sqrt{10}}{3}...\left(2\right) \end{gathered}$$
Now,
$$\begin{gathered} 3x^2+4xy-3y^2 \\ =3\left(x^2-y^2\right)+4xy \\ =3\left(x+y\right)\left(x-y\right)+4xy \\ =3\left(\frac{7+2\sqrt{10}}{3}+\frac{7-2\sqrt{10}}{3}\right)\left(\frac{7+2\sqrt{10}}{3}-\frac{7-2\sqrt{10}}{3}\right)+4\left(\frac{7+2\sqrt{10}}{3}\times \frac{7-2\sqrt{10}}{3}\right) \\ =3\left(\frac{14}{3}\right)\left(\frac{4\sqrt{10}}{3}\right)+4\left(\frac{{\left(7\right)}^2-{\left(2\sqrt{10}\right)}^2}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4\left(\frac{49-40}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4 \end{gathered}$$
Hence, $3x^2+4xy-3y^2=4+\frac{56\sqrt{10}}{3}$.